1. Stating the problem: Solve the equation $$(2x^2 - 3x)^2 - 2x^2 + 3x = 2.$$\n\n2. Expand the square term: $$(2x^2 - 3x)^2 = (2x^2)^2 - 2\times 2x^2 \times 3x + (3x)^2 = 4x^4 - 12x^3 + 9x^2.$$\n\n3. Substitute back into the equation: $$4x^4 - 12x^3 + 9x^2 - 2x^2 + 3x = 2.$$\n\n4. Simplify the quadratic terms: $$4x^4 - 12x^3 + (9x^2 - 2x^2) + 3x = 2,$$ so $$4x^4 - 12x^3 + 7x^2 + 3x = 2.$$\n\n5. Bring all terms to one side for standard polynomial form: $$4x^4 - 12x^3 + 7x^2 + 3x - 2 = 0.$$\n\n6. The equation to solve is: $$4x^4 - 12x^3 + 7x^2 + 3x - 2 = 0.$$\n\n7. Solve this quartic equation. Try rational root test. Possible rational roots are factors of 2 over factors of 4: $$\pm1, \pm\frac{1}{2}, \pm2, \pm\frac{1}{4}.$$\n\n8. Test $x=1$: $$4(1)^4 -12(1)^3 + 7(1)^2 + 3(1) - 2 = 4 -12 +7 +3 - 2 = 0.$$ So $x=1$ is a root.\n\n9. Use polynomial division or synthetic division to divide by $(x-1)$: $$4x^4 - 12x^3 + 7x^2 + 3x - 2 \div (x-1) = 4x^3 - 8x^2 - x + 2.$$\n\n10. Now solve the cubic equation: $$4x^3 - 8x^2 - x + 2 = 0.$$\n\n11. Try rational root test again for the cubic: possible roots $\pm1, \pm\frac{1}{2}, \pm2, \pm \frac{1}{4}.$\n\n12. Test $x=1$: $$4(1) -8(1) -1 + 2 = 4 -8 -1 + 2 = -3 \neq 0.$$\n\n13. Test $x=2$: $$4(8) -8(4) -2 + 2 = 32 -32 -2 + 2 = 0.$$ So $x=2$ is a root.\n\n14. Divide cubic by $(x-2)$: quotient is $$4x^2 - 0x - 1 = 4x^2 - 1.$$\n\n15. Now solve quadratic: $$4x^2 - 1 = 0.$$\n\n16. Solve quadratic: $$4x^2 = 1 \implies x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2}.$$\n\n17. Solutions are $$x = 1, 2, \frac{1}{2}, -\frac{1}{2}.$$\n\n18. Final answer: $$\boxed{x = -\frac{1}{2}, \frac{1}{2}, 1, 2}.$$